Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
maybe u had to multiply the rate and monthly payment
Answer:
A line passes (-7, 0), (-7, -6).
These two passed points have same x-component (x = -7).
=> This is the line which has equation: x = -7.
This is a vertical line (parallel with y axis) and the slope is undefined.
Hope this helps!
:)
Positive 3, If You're starting at -6 and you add the first 6 it's at 0, now you have 3 more to add to make 9. And that gives you positive 3. or if you do 9 - 6 you can get your answer that way too.